Question

Consider an experiment whose sample space consists of a countably infinite number of points. Show that not all points can be equally likely. Can all points have a positive probability of occurring?

Answer 1

Yes they do, if they had a negative propabaility then the points would have a chance of not occuring.

Answer 2

Where 𝜆=5, x=0: P(x)=(e^(-5) 5^0)/0! =e-5=0.006738
Where 𝜆=5, x=1: P(1)= (e^(-5) 5^1)/1! =5e-5=0.0337
Where 𝜆=5, x=2: P(2)= (e^(-5) 5^2)/2! =25e-5=0.0842
Thus, not all points are equally likely, but we can say that all points have a positive probability of occurring.

Answer 3

How do I show this though?
I think whatever I did is wrong

Answer 4

It's question 4 in the attached file if whatever I wrote is coming up weird characters

Answer 5

@swagmaster47

Answer 6

I'm not sure if this consists as enough proof.. but if your distribution assumes that all occurrences are equally likely, then \(f(x)=\dfrac{1}{x}, x >0\) But the sum of all probabilities must be 1, i.e. \(P(\Omega)=1\)

But \[\sum_{x=1}^{\infty}\frac{1}{x} \] is the divergent harmonic series.

Answer 7

Hm I should have written the suport above as \(x=1,2,3,...\) not \(x>0\).

As for all events having positive probabilities.. yes that certainly can happen. Well it wouldn't be much of a probability distribution if the probabilities could be negative...

An example would be the geometric distribution.
\[ f(x)=(1-p)^{x-1}p, x=1,2,3...\]
\(f(x)>0 \) for all \(x\)

Answer 8

For your example, it looks like the Poisson distribution. The support for the Poisson is \(x \in \{0,1,2,... \}\)
And you can show that all these probabilities are positive since
\[f(x)=\frac{e^{-\lambda}\lambda^x}{x!} \] is always positive since, \(e^{-x}>0\) for all x, \(\lambda^x>0\) for all x, because \(\lambda>0\).

And dividiving this result by \(x!\) ensures it's positive, because \(0!, 1!, 2!, 3!, ...\) are all \(>0\).

Answer 9

Check out my updated file?

Answer 10

I guess it looks correct.